Question: What is the degree measure of angle $LOQ$ when polygon $\allowbreak LMNOPQ$ is a regular hexagon? [asy]
draw((-2,0)--(-1,1.73205081)--(1,1.73205081)--(2,0)--(1,-1.73205081)--(-1,-1.73205081)--cycle);
draw((-1,-1.73205081)--(1,1.73205081)--(1,-1.73205081)--cycle);
label("L",(-1,-1.73205081),SW);
label("M",(-2,0),W);
label("N",(-1,1.73205081),NW);
label("O",(1,1.73205081),N);
label("P",(2,0),E);
label("Q",(1,-1.73205081),S);
[/asy]
The sum of the interior angles of an $n$-sided polygon is $180(n-2)$. For a regular hexagon, the interior angles sum to $180(4)$, so each interior angle has a measure of $\frac{180(4)}{6}=30\cdot4=120^\circ$. Since $\overline{PO}$ and $\overline{PQ}$ are congruent sides of a regular hexagon, $\triangle POQ$ is an isosceles triangle. The two base angles are congruent and sum to a degree measure of $180-120=60^\circ$, so each base angle has a measure of $30^\circ$. There are now a couple approaches to finishing the problem.

$\emph{Approach 1}$: We use the fact that trapezoid $PQLO$ is an isosceles trapezoid to solve for $x$ and $y$. Since $\overline{PO}$ and $\overline{QL}$ are congruent sides of a regular hexagon, trapezoid $PQLO$ is an isosceles trapezoid and the base angles are equal. So we know that $x+30=y$. Since the interior angle of a hexagon is $120^\circ$ and $m\angle PQO=30^\circ$, we know that $\angle OQL$ is a right angle. The acute angles of a right triangle sum to $90^\circ$, so $x+y=90$. Now we can solve for $x$ with $x+(x+30)=90$, which yields $x=30$. The degree measure of $\angle LOQ$ is $\boxed{30^\circ}$.

$\emph{Approach 2}$: We use the fact that trapezoid $LMNO$ is an isosceles trapezoid to solve for $x$. Since $\overline{NO}$ and $\overline{ML}$ are congruent sides of a regular hexagon, trapezoid $LMNO$ is an isosceles trapezoid and the base angles are equal. The interior angles of a trapezoid sum to $360^\circ$, so we have $2z+120+120=360$, which yields $z=60$. Angle $O$ is an interior angle of a hexagon that measure $120^\circ$, so $z+x+30=120$. We found that $z=60$, so $x=30$. The degree measure of $\angle LOQ$ is $\boxed{30^\circ}$.

[asy]
pen sm=fontsize(9);
draw((-2,0)--(-1,1.73205081)--(1,1.73205081)--(2,0)--(1,-1.73205081)--(-1,-1.73205081)--cycle);
draw((-1,-1.73205081)--(1,1.73205081)--(1,-1.73205081)--cycle);
label("L",(-1,-1.73205081),SW);
label("M",(-2,0),W);
label("N",(-1,1.73205081),NW);
label("O",(1,1.73205081),N);
label("P",(2,0),E);
label("Q",(1,-1.73205081),S);
label("$120^\circ$", (2,0), W, sm);
label("$120^\circ$", (-2,0), E, sm);
label("$120^\circ$", (-1,1.73205081), SE, sm);
label("$30^\circ$", (1,0.93205081), SE, sm);
label("$x^\circ$", (0.8,1.53205081)-(0,0.2), S, sm);
label("$z^\circ$", (0.9,1.73205081), SW, sm);
label("$30^\circ$", (1,-0.93205081), NE, sm);
pair O=(1,1.73205081), Q=(1,-1.73205081), L=(-1,-1.73205081);
label("$y^\circ$", L+(0.1,0.1), ENE, sm);
label("$z^\circ$", L+(0,0.2), N, sm);
draw(rightanglemark(O,Q,L));
[/asy]